1. **State the problem:** Calculate the enthalpy change $\Delta H$ for the reaction: $$\mathrm{CO_{(g)} + NO_{(g)} \rightarrow CO_2{(g)} + \frac{1}{2} N_2{(g)}}$$ 2. **Given reactions and their enthalpy changes:** - Equation A: $$\mathrm{CO_{(g)} + \frac{1}{2} O_2{(g)} \rightarrow CO_2{(g)}} \quad \Delta H_A = -383.0$$ - Equation B: $$\mathrm{N_2{(g)} + O_2{(g)} \rightarrow 2 NO_{(g)}} \quad \Delta H_B = 280.6$$ 3. **Goal:** Use Hess's Law, which states that the enthalpy change of a reaction is the sum of enthalpy changes of steps that lead to it. 4. **Manipulate given equations to match the target reaction:** - Reverse Equation B to get $$\mathrm{2 NO_{(g)} \rightarrow N_2{(g)} + O_2{(g)}}$$ which changes the sign of $\Delta H_B$: $$\Delta H = -280.6$$ - Divide this reversed equation by 2 to get: $$\mathrm{NO_{(g)} \rightarrow \frac{1}{2} N_2{(g)} + \frac{1}{2} O_2{(g)}}$$ with enthalpy change: $$\Delta H = \frac{-280.6}{2} = -140.3$$ 5. **Add Equation A and the modified Equation B:** $$\mathrm{CO_{(g)} + \frac{1}{2} O_2{(g)} \rightarrow CO_2{(g)}} \quad (\Delta H = -383.0)$$ $$\mathrm{NO_{(g)} \rightarrow \frac{1}{2} N_2{(g)} + \frac{1}{2} O_2{(g)}} \quad (\Delta H = -140.3)$$ Adding these: $$\mathrm{CO_{(g)} + NO_{(g)} + \frac{1}{2} O_2{(g)} \rightarrow CO_2{(g)} + \frac{1}{2} N_2{(g)} + \frac{1}{2} O_2{(g)}}$$ Cancel $\frac{1}{2} O_2{(g)}$ on both sides: $$\mathrm{CO_{(g)} + NO_{(g)} \rightarrow CO_2{(g)} + \frac{1}{2} N_2{(g)}}$$ 6. **Calculate total enthalpy change:** $$\Delta H = -383.0 + (-140.3) = -523.3$$ **Final answer:** $$\boxed{\Delta H = -523.3}$$ This means the reaction releases 523.3 units of enthalpy (kJ) and is exothermic.